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Linear-Algebra · Eigenvalues

Eigenvalues of [[3,1],[1,3]]

\left[\begin{matrix}3 & 1\\1 & 3\end{matrix}\right]

Step by step

  1. \det(A - \lambda I) = 0

    Eigenvalues are the roots of the characteristic polynomial.

  2. \det\left[\begin{matrix}3 - \lambda & 1\\1 & 3 - \lambda\end{matrix}\right] = 0

    Subtract λ from the diagonal.

  3. \lambda^{2} - 6 \lambda + 8 = 0

    Expand the determinant.

  4. \left(\lambda - 4\right) \left(\lambda - 2\right) = 0

    Factor.

  5. \lambda = 4, \lambda = 2

    Eigenvalues (with multiplicity).

  6. \lambda = 2:\ (A - \lambda I)\mathbf{v} = 0 \Rightarrow \mathbf{v} = \left[\begin{matrix}-1\\1\end{matrix}\right]

    Solve (A − 2I)v = 0 for a basis eigenvector.

  7. \lambda = 4:\ (A - \lambda I)\mathbf{v} = 0 \Rightarrow \mathbf{v} = \left[\begin{matrix}1\\1\end{matrix}\right]

    Solve (A − 4I)v = 0 for a basis eigenvector.

Reveal the answer
\lambda = 4,\; \lambda = 2